As explained in Part I of the article, instead of building their house with the easy method of hexagonal top + hexagonal bottom, honeybees are building their cells with hexagonal top + rhombic bottom. In this article, I will try to give a calculus-based demonstration on how the honeybees achieve the economization of their wax resources with a rhombic angle of 70°32'.
Suppose we start with a hexagonal prism with flat bottom like the one shown in the figure above. Let \( AB = a\), \(A''A = b\). Then it can be shown that the surface area \(S_0\) and the volume \(V\) of this prism are:
$$S_0= 6ab + \frac{3\sqrt{3}}{2}a^2, \quad V = \frac{3\sqrt{3}}{2}a^2b$$
respectively. Mark a point \(B'\) on the prism so that \(B'B'' = x\) and slice a tetrahedron from the prism along the line \(A''C''\) through point \(B'\). Then, flip and rearrange the sliced tetrahedron on top of the prism, as shown below.
If you repeat this procedure for the other two sides of the prism, you will end up with a prism shown below.
Since we are doing slicing and rearrangement of parts on the prism, the volume of this rearranged prism is conserved. However, the surface area \(S_1\) of this reformed prism no longer the same as \(S_0\). It is not difficult to show that the \(S_1\) is:
$$S_1 = 6ab - 3ax +3a\sqrt{3} \sqrt{x^2 + \tfrac{1}{4}a^2}$$
The difference between the two surface areas (i.e. the amount of wax saved) is thus:
$$\Delta S(x) = S_0 - S_1 = \frac{3\sqrt{3}}{2}a^2 + 3ax - 3\sqrt{3}a\sqrt{x^2 + \tfrac{1}{4}a^2}$$
To minimize the amount of wax used, the honeybees must maximize the value of \(\Delta S\). This is achieved by setting the first derivative of \(\Delta S\) to zero:
$$\Delta S'(x) = 3a - 3\sqrt{3}ax \left(x^2 + \tfrac{1}{4}a^2\right)^{-1/2} = 0$$
The solution to this equation is:
$$x = \tfrac{1}{\sqrt{8}}a$$
Since \(A'A'' = \frac{1}{\sqrt{8}}a\) and \(A'B = a\), the hypotenuse \(A''B'\) must be \(\frac{3}{\sqrt{8}}a\). For a regular hexagon, we have \(A''C'' = a\sqrt{3}\), and therefore the rhombic angle \(GA''B'\) must be $$2 \cos^{-1}\sqrt{\tfrac{2}{3}} = 70.5288 = 70^\circ 32'$$
To understand why Nature is a stingy mathematician. See Part III of this blog series.
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Anonymous said…
Can you please Email me the work showing how x=a/(8^1/2) is substituted into honeycomb cell geometry to arrive at the rhombic angles. Email me at jdelatorre@dusd.net Thank you!
Since A'A'' = a/8^0.5 and A'B = a, hypotenuse A''B' is 3a/8^0.5. For a regular hexagon, we have A''C" = a*3^0.5. Now the angle GA"B' be written in term of cosine as 2*acos((2/3)^0.5) = 70.529 degrees
如果你或你的公司不還稅， 內陸稅收局 的人就會寄一封公函來罵你並跟你討錢。 在馬來西亞，所有的政府公函，都會印上 Urusan Seri Paduka Baginda 的字樣。「Urusan Seri Paduka Baginda」其實是延續英國人 On His Majesty's Service 的傳統。「 Seri Paduka Baginda 」這三個字的英語是「His Majesty」或「Her Majesty」，而「Urusan」則是「Service」。 現代英國郵遞系統的原型是愛德華四世國王 Edward IV (1461-1483)設計的，他為了加快情報的傳遞速度，在大概每30km就設了一根柱子(post)和一匹新馬。這樣，情報的傳遞便能達到約2 m/s。 J. C. Hemmeon在1912年出版的《The history of the British post office》是這樣寫道的： A somewhat different style of postal service, a precursor of the modern method, was inaugurated by the fourth Edward. During the war with Scotland he found himself in need of a speedier and better system of communication between the seat of war and the seat of government. He accomplished this by placing horses at intervals of twenty miles along the great road between England and Scotland. By so doing his messengers were able to take up fresh horses along the way and his despatches were carried at the rate of a hundred miles a day. 500年前2 m
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